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        Transformer家族之Non-Monotonic Transformer
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          <p><img src="https://cdn.jsdelivr.net/gh/rogerspy/blog-imgs/5396ee05ly1g5pqn3ch6zj20u092znph.jpg" alt></p>
<p>之前我们介绍的两种 <em>insertion-based</em> 文本生成方法预先规定了每次生成最中间的词，这样一来我们虽然利用树实现了并行，但是却丢失了其中的生成模式，我们不知道模型在生成的时候经历了什么。那么我们能不能让模型自动生成一棵树呢？比如，现在生成了一个根节点，然后再生成左右子节点，然后再生成子节点的子节点，以此类推，但不同的是，这棵树不一定平衡，甚至可能退化成一条链，但我们获得了模型的生成模式，如下图所示：</p>
<a id="more"></a>
<p><img src="https://cdn.jsdelivr.net/gh/rogerspy/blog-imgs/20200421163414.png" alt></p>
<ul>
<li>可以从任意位置开始生成；</li>
<li>图中绿框中的数字表示生成顺序；</li>
<li>图中蓝筐中 的数字表示重构顺序；</li>
<li>传统的从左向右被做为二叉树的一种特殊情况。</li>
</ul>
<h1 id="1-Abstract-Framework"><a href="#1-Abstract-Framework" class="headerlink" title="1. Abstract Framework"></a>1. Abstract Framework</h1><ul>
<li>$Y=(w_1, …, w_N)$ ——表示生成的序列，其中 $w_i \in V$，是一个词表;</li>
<li>$\tilde{V} = V \bigcup \{&lt; end &gt;\}$ —— 表示所有可能生成的字符串；</li>
<li>$S = \tilde{V}^*$ —— 表示序列的状态空间；</li>
<li>$s \in S$ —— 表示状态空间中的一个序列，该序列中每个元素都来源于 $\tilde{V}$。比如上图 $s_1=(are), s_2=(are, how),…,s_4=(are, how, ?, &lt; end &gt;)$；</li>
<li>$a$ —— 表示将 $\tilde{V}$ 中的一个元素添加到 $S$ 后面的操作；</li>
<li>所有的叶子结点都是 $&lt; end &gt;$ 时表示序列生成过程的结束，此时 $T=2N+1$，其中 $N$ 表示序列中非 $&lt; end &gt;$ 的元素， $T$ 表示整个序列的长度， $s_T$ 表示最终状态。</li>
<li>$\tau(t)$ —— 表示按层序遍历树上的第 $t$ 的节点，则 $(a_{\tau(1)},…,a_{\tau(T)})$；</li>
<li>以上图为例，最终的序列是将图中蓝筐中的数字映射成绿框的顺序，然后删掉所有的 $&lt; end &gt;$。</li>
<li>$\pi(a|s)$ —— 表示根据给定一个序列状态生成一个 $a$ 操作的策略。我们知道树的遍历有很多种方式，因此对于同一个序列，我们可以有不同的树，我们就有多种生成模式。</li>
</ul>
<h1 id="2-Learning-for-Non-Monotonic-Generation"><a href="#2-Learning-for-Non-Monotonic-Generation" class="headerlink" title="2. Learning for Non-Monotonic Generation"></a>2. Learning for Non-Monotonic Generation</h1><p>我们考虑两种情况下的文本生成：</p>
<ul>
<li>非条件生成：类似语言模型，没有额外信息输入；</li>
<li>条件生成：类似机器翻译，根据输入序列生成新的序列。</li>
</ul>
<p>我们先考虑非条件生成的情况。这种情况比较复杂，因为我们只知道最终的序列是什么样的，但是生成这样的序列的那棵树长什么样子我们并不知道。因为我们是通过遍历树的叶子结点得到最终的序列，而树的遍历有很多种方法，要用不同的遍历方法得到同一个序列，此时树的结构就不尽相同。树的结构不同意味着每次生成都会有多种操作的可能性，这样我们就不能使用传统的监督学习方法了。为了解决这个问题我哦们可以使用 <em>learning-to-search</em> 和 <em>imitation learning</em>。</p>
<blockquote>
<p><strong>Key  Idea</strong></p>
<p>假定我们只有一个序列 $Y$。现在的想法是，在第一步我们首先生成任意一个单词 $w \in Y$ 作为整棵树的根节点，然后类似快速排序的思想，在 $w$ 左边和右边递归地生成，由于我们希望树的中序遍历可以得到原始序列 $Y$，所以 $w$ 左边的字符必须在 $w$ 的左子树，同理对右子树。然后用 <em>direct loss minimization</em> 及相关的技术学习一个参考策略 $\pi^{\ast}$ 当成当前策略 $\pi $ 的首选策略。</p>
</blockquote>
<h2 id="2-1-Unconditional-Generation"><a href="#2-1-Unconditional-Generation" class="headerlink" title="2.1 Unconditional Generation"></a>2.1 Unconditional Generation</h2><p>所谓 <em>Learning-to-search</em>，就是模仿一个参考策略 $\pi^{\ast}$ 来学习当前策略 $\pi$。我们定义一个输入策略（<em>roll-in</em>）$\pi^{in}$ 和一个输出策略（<em>roll-out</em>）$\pi^{out}$。下面我们不断从 $\pi^{in}$ 中抽样状态 $s$，然后在 $\pi^{out}$ 下对每个行为 $a$ 计算一个运行代价，之后学到的 $\pi $ 被训练去最小化这个运行代价。</p>
<p>用数学语言来说就是:</p>
<script type="math/tex; mode=display">
\mathbb{E}_{Y \sim D}\mathbb{E}_{t \sim U([1, 2|Y|+1])}\mathbb{E}_{s_t\sim d_{\pi^{in}}^t}\left[C(\pi;\pi^{out}, s_t) \right]</script><ul>
<li>$U(T)$ 表示 $[1,…,T]$ 的均匀分布；</li>
<li>$d_\pi^{in}$ 表示在 $\pi$ 策略下进行 $t$ 步得到的状态分布；</li>
<li>$C(\pi;\pi^{out}, s)$ 表示运行代价</li>
</ul>
<p>通过选择不同的 $\pi;\pi^{out}, s$ 我们可以得到不同的 <em>learning-to-search</em> 算法，我们希望找到一个策略能够在获得 $s_t$ 上表现得和 $\pi^*$ 一样好，甚至更好。</p>
<ul>
<li><p><strong>$\pi^{in}$ 的选择</strong></p>
<p>$\pi^{in}$ 决定了我们要学习的策略 $\pi$ 训练的状态分布。我们可以选择 $\pi$ 与 $\pi^{out}$ 的混合作为 $\pi^{in}$，也可以选择仅使用 $\pi^{out}$ 。后者更简单，我们本着从简的原则，使用后者。</p>
</li>
<li><p><strong>$\pi^{out}$ 的选择</strong></p>
<p>$\pi^{out}$ 就是我们要模仿的策略。由于 $\pi^{out}$ 是参考策略，所以我们可以通过 $\pi ^{out}$ 完全根据序列构建出树，于是我们可以把 $\pi^{out} $ 视为树的生成过程，在每一步都对应一个状态 $s_t$ 和序列 $Y_t$ 。在每个 $s_t$ ，$Y_t$ 包含了合法行为，比如下图，在第一步，我们可以选择生成 $(a, b, c, d)$ 中的一个，比如我们选择了 $b$，之后，左节点的选择就只有 $a$ 了，右节点的选择就有 $(c,d)$，这些选择就是“合法”的。</p>
<p><img src="https://cdn.jsdelivr.net/gh/rogerspy/blog-imgs/20200422192321.png" alt></p>
<p>给定一个连续子序列 $Y_t=(w_1’, …, w_{N^{‘}}’)$，$\pi^{out}$ 可以定义为：</p>
<p><img src="https://cdn.jsdelivr.net/gh/rogerspy/blog-imgs/20200422193519.png" alt></p>
<p>其中 $\sum_{a \in Y}p_a=1$。我们可以看到 $\pi^{out}$ 策略主要由 $p_a$ 决定，这里作者预定义了三种不同的策略：均匀策略（或者任意顺序策略，<em>uniform oracle</em>）、指导策略（<em>coaching oracle</em>）和退火指导策略（<em>annealed coaching oracle</em>）。</p>
<ul>
<li><p><strong>均匀策略</strong></p>
<p>令 $p_a = 1/n$，记为 $\pi^{\ast}_{\mathrm{uniform}}$。</p>
</li>
<li><p><strong>指导策略</strong></p>
<p>任意顺序的策略会导致一个问题：难以使得 $\pi $ 去模仿。为此，我们可以考虑加入当前学习的策略 $\pi  $：</p>
<script type="math/tex; mode=display">
\pi^{\ast}_{\mathrm{coaching}}(a|s) \propto \pi^*_{\mathrm{uniform}}(a|s)\pi(a|s)</script><p>这样一来，既可以避免不合法的行为，也可以按照当前策略 $\pi $ 继续学习。</p>
</li>
<li><p><strong>退火指导策略</strong></p>
<p>指导策略也有一个问题：它不会引导模型进行多样化的学习。因此，我们可以再适当加入$\pi^{\ast}_{\mathrm{uniform}}$：</p>
<script type="math/tex; mode=display">
\pi^*_{\mathrm{annealed}}(a|s) = \beta \pi^{\ast}_{\mathrm{uniform}}(a|s) | (1-\beta) \pi^{\ast}_{\mathrm{coaching}}(a|s)</script><p>这里 $\beta$ 随着训练从1线性递减到0。</p>
</li>
</ul>
</li>
<li><p><strong>$C$ 的选择</strong></p>
<p>$C$ 度量的是经过输入策略的选择得到的状态与输出策略之间的误差，最常见的方法是使用平方误差。然而，有研究表明使用 <em>RNN</em> 时平方误差表现不佳，所以我们可以转而使用 <em>KL</em> 散度:</p>
<script type="math/tex; mode=display">
C(\pi, \pi^{out}, s) = D_{KL}(\pi^{out}(\cdot|s)||\pi(\cdot|s)) = \sum_{a \in \tilde{V}}\pi^{out}(a|s)\log \pi(a|s)+C</script></li>
</ul>
<h2 id="2-2-Conditional-Generation"><a href="#2-2-Conditional-Generation" class="headerlink" title="2.2 Conditional Generation"></a>2.2 Conditional Generation</h2><p>上面说的是给定单个句子 $Y$ ，如果我们要学习 $X \rightarrow Y$ 怎么办呢？</p>
<p>我们将条件输入 $X$ 编码成一组 $d_{enc}$ 维的向量，记为 $f^{enc}(X)$，然后经过神经网络，比如 <em>LSTM</em> 或者 <em>Transformer</em>，得到隐状态向量 $H \in \mathbb{R}^{|X| \times d_{enc}}$，然后再送入该模型中即可。</p>
<h1 id="3-神经网络结构"><a href="#3-神经网络结构" class="headerlink" title="3. 神经网络结构"></a>3. 神经网络结构</h1><p>作者选择使用神经网络实现上面的二叉树生成策略，因为神经网络能有效的对不同尺寸的输入进行编码以及预测输出。这里作者选择两种神经网络：<em>LSTM</em> 和 <em>Transformer</em>，一个是老一代的 <em>NLP</em> 武林盟主，一个是新生代江湖俊杰。</p>
<h2 id="3-1-LSTM-Policy"><a href="#3-1-LSTM-Policy" class="headerlink" title="3.1 LSTM Policy"></a>3.1 LSTM Policy</h2><p>将二叉树的层序遍历节点 $(a_1, …, a_t)$ 作为序列输入到 <em>LSTM</em> ，然后 <em>LSTM</em> 将序列编码成向量 $h_t$，然后计算 $a_i$ 的概率分布：</p>
<script type="math/tex; mode=display">
\pi(a|s_t) \propto \exp(u_a^Th_t+b_a)</script><h2 id="3-2-Transformer-Policy"><a href="#3-2-Transformer-Policy" class="headerlink" title="3.2 Transformer Policy"></a>3.2 Transformer Policy</h2><p>同样是将 $(a_1, …, a_t)$ 作为输入，然后使用多头注意力计算 $h_t$，再计算 $a_i$ 的概率分布。</p>
<h2 id="3-3-Auxiliary-lt-end-gt-Prediction"><a href="#3-3-Auxiliary-lt-end-gt-Prediction" class="headerlink" title="3.3 Auxiliary $&lt; end &gt;$ Prediction"></a>3.3 Auxiliary $&lt; end &gt;$ Prediction</h2><p>作者将 $&lt; end &gt;$ 的预测和 $a_i$ 的预测分开进行，先利用伯努利分布判断是否是 $&lt; end &gt;$，设定一个阈值 $\tau$ ，当概率大于 $\tau$ 时，则认为 $a_t = &lt; end &gt;$，否则根据 $\pi$ 计算 $a_t$。</p>
<h1 id="4-Experiments"><a href="#4-Experiments" class="headerlink" title="4. Experiments"></a>4. Experiments</h1><p>作者最后在多个任务上进行了实验：</p>
<ol>
<li>语言模型</li>
<li>句子补齐</li>
<li>词序重排</li>
<li>机器翻译</li>
</ol>
<p>这里展示几个生成样例。</p>
<ul>
<li>语言模型</li>
</ul>
<p><img src="https://cdn.jsdelivr.net/gh/rogerspy/blog-imgs/20200423101930.png" alt></p>
<ul>
<li>机器翻译</li>
</ul>
<p><img src="https://cdn.jsdelivr.net/gh/rogerspy/blog-imgs/20200423110508.png" alt></p>
<p>从图中我们可以清晰地看到生成模式。更多详细的实验可以去看原文。</p>
<h1 id="Reference"><a href="#Reference" class="headerlink" title="Reference"></a>Reference</h1><ol>
<li><a href="https://arxiv.org/pdf/1902.02192.pdf" target="_blank" rel="noopener">Non-Monotonic Sequential Text Generation</a>, <em>Sean Welleck, Kiante Brantley, Hal Daume III, Kyunghyun Cho. 2019. arXiv: 1902.02192</em></li>
<li><a href="https://github.com/kweonwooj/papers/issues/121" target="_blank" rel="noopener">Non-Monotonic Sequential Text Generation #121</a>, <em>kweonwooj. Github Pages. 2019</em></li>
<li><a href="https://zhuanlan.zhihu.com/p/73417154" target="_blank" rel="noopener">香侬读 | 按什么套路生成？基于插入和删除的序列生成方法</a>, 香侬科技，知乎</li>
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之前我们介绍的两种 insertion-based 文本生成方法预先规定了每次生成最中间的词，这样一来我们虽然利用树实现了并行，但是却丢失了其中的生成模式，我们不知道模型在生成的时候经历了什么。那么我们能不能让模型自动生成一棵树呢？比如，现在生成了一个根节点，然后再生成左右子节点，然后再生成子节点的子节点，以此类推，但不同的是，这棵树不一定平衡，甚至可能退化成一条链，但我们获得了模型的生成模式，如下图所示："
          
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之前我们介绍的两种 insertion-based 文本生成方法预先规定了每次生成最中间的词，这样一来我们虽然利用树实现了并行，但是却丢失了其中的生成模式，我们不知道模型在生成的时候经历了什么。那么我们能不能让模型自动生成一棵树呢？比如，现在生成了一个根节点，然后再生成左右子节点，然后再生成子节点的子节点，以此类推，但不同的是，这棵树不一定平衡，甚至可能退化成一条链，但我们获得了模型的生成模式，如下图所示："
          
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之前我们介绍的两种 insertion-based 文本生成方法预先规定了每次生成最中间的词，这样一来我们虽然利用树实现了并行，但是却丢失了其中的生成模式，我们不知道模型在生成的时候经历了什么。那么我们能不能让模型自动生成一棵树呢？比如，现在生成了一个根节点，然后再生成左右子节点，然后再生成子节点的子节点，以此类推，但不同的是，这棵树不一定平衡，甚至可能退化成一条链，但我们获得了模型的生成模式，如下图所示："
          
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